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\title{Introduction to Quantum Mechanics\\Lecture 5}
\author{Brian Greene\footnote{These lecture notes were TeX'd by Nilay Kumar and Carlos Hernandez}} % ADD YOUR NAME HERE IF YOU WORK ON THIS LECTURE's NOTES
\date{September 27, 2011}

\maketitle

\section{Momentum space}

Recall that for a free particle ($V=0$) with momentum $p=\hbar k$ the wavefunction is of the form $e^{ikx},$
which is sinusoidal with wavenumber $k$. We can use Fourier analysis to write any wavefunction $\psi(x)$ as a linear combination of
sinusoids, and thus as a linear combination of free particle wavefunctions with arbitrary momenta:
\begin{equation}
    \begin{split}
    \psi(x)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\\
    \text{with } \phi(k)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\psi(x)e^{-ikx}\,dx
\end{split}
    \label{eq:fourier}
\end{equation}
This motivates us to think of $\psi(x)$ as the wavefunction of a particle that has, in a sense, \textit{more than one} momentum, an idea
that is in stark contrast with the ideas of classical physics. Similarly, we can think of $\phi(k)$ to be weights applied to the free
particle wavefunctions. More precisely, $|\phi(k)|^2$ represents ``how much'' of the momentum $p=\hbar k$ is present in $\psi(x)$.

We now have a very interesting result! Just as $|\psi(x)|^2\,dx$ gives us the probability of the particle being located at $x$,
$|\phi(k)|^2\,dk$ gives us the probability of the particle's momentum being $\hbar k$. In other words, $\psi(x)$ is the particle's
wavefunction in $x$-space, whereas $\phi(k)$ is the particle's wavefunction in $k$-space.

\section{Uncertainty}
\subsection{Position and Momentum}
Here, we will  introduce the concept of uncertainty as it pertains to a particle's position and momentum. Suppose that we want to measure the position and speed of an object--- for example a baseball being hit into center field. Naively, we assume that the baseball has a definite position and speed at a particular moment in time, and how accurately we can measure these values depends on the quality of our measuring equipment--- if we improve the precision of our measuring equipment, we will get a result that is closer to the true value. In particular, we would assume that how precisely we measure the speed of the baseball does not affect its trajectory, and vice versa.

\begin{figure}[h!]

\centering
\href{http://www.columbia.edu/~cxh2102/qm/uncertainty.swf}{
\includegraphics[width=1\textwidth]{uncertainty.jpg}}
\caption{A brief demonstration on how increasing the uncertainty of knowing a particle's position in space can increase one's certainty in knowing its momentum. Please click the graphic to begin.}
\end{figure}

In 1927, Heisenberg proved that these assumptions are not correct. Quantum mechanics shows that position and speed of an object cannot both be known to arbitrary precision: the more precisely one property is known, the less precisely the other can be known. This statement is known as the uncertainty principle. This is not a statement about the accuracy of our measuring equipment, but about the nature of the system itself--- our naive assumption that the baseball had a definite position and speed was incorrect. On a scale of people and baseballs, these uncertainties are too small to notice, but when dealing with atoms and electrons they become crucial.
The uncertainty principle shows

\begin{equation}
\Delta x\Delta p\geq\frac{\hbar }{2}
\end{equation}

where $\Delta x$ and $\Delta p$ represent the error (or uncertainty) of an object's position and momentum, respectively.

\subsection{Why Don't I Notice Quantum Effects?}
Given what we've learned thus far, conceivably, objects could travel unpredictably through space due to their intrinsically uncertain momentum--- however we know empirically that this is certainly not true in our everyday lives. Let us continue to use the example of a baseball, given the following assumptions, to explain why quantum effects are insignificant at our macroscopic scale.
\begin{equation*}
\label{eq:uncertainty}
\begin{array}{l l}
  m = 150 \text{ g}\\
 \Delta x = 0.1 \text{ mm}\\
  \end{array}
\end{equation*}
where \emph{m} is the mass of the baseball and $\Delta x$ represents how accurately we can pinpoint where the baseball is located in space--- in this case, about the thickness of a human hair. 

Now let us plug these values into Eq.~(\ref{eq:uncertainty}) to find the intrinsic speed of the baseball due to quantum effects
\begin{equation*}
\begin{array}{l l}

  \Delta v = \frac{\hbar }{2 m \Delta x} = \frac{(10^{-34} \text{ J s}) }{2 (0.15 \text{ kg}) (10^{-4} \text{ m})} \approx 4 \times 10^{-30}  \text{ m/s}
  \end{array}
\end{equation*}

This is incredibly slow! According to the uncertainty principle, the time it would take the baseball to move just an angstrom (the length scale of an atom) due to quantum effects would be roughly sixty times the age of the universe.




\section{The Dirac delta function}

We are now going to explore the properties of the Dirac delta function\footnote{Strictly speaking, the Dirac delta is a distribution (or a
generalized function) and not an actual function, but the distinction is inconsequential to our purposes here.}, which is an especially
useful mathematical tool.
\subsection{Derivation}
 Starting with the equations for the Fourier transform, we have
\begin{equation*}
    \begin{split}
    \psi(x)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\\
    \text{with } \phi(k)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\psi(x')e^{-ikx'}\,dx'
\end{split}
\end{equation*}
If we insert the second equation into the first, we get the following double integral:
\begin{equation}
    \psi(x)=\frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty\psi(x')e^{ik(x-x')}\,dk\,dx'
    \label{eq:fouriertodelta}
\end{equation}
If we define the Dirac delta function to be\footnote{We write $\delta(x-x')$ instead of $\delta(x,x')$ because all that matters is the
difference $x-x'$ - neither $x$ nor $x'$ is used independently in the defintion.}
\begin{equation}
    \delta(x-x')=\frac{1}{2\pi}\int_{-\infty}^\infty e^{ik(x-x')}\,dk,
    \label{eq:delta}
\end{equation}
we can rewrite Eq.~(\ref{eq:fouriertodelta}) as
\begin{equation*}
    \psi(x)=\int_{-\infty}^\infty\psi(x')\delta(x-x')\,dx'.
\end{equation*}
So somehow, when placed under an integral with respect to $x$, the delta function $\delta(x-x')$ ``picks out'' the value of the integrand at
$x=x'$.

More generally, we find that Dirac delta function behaves as a collapsed Gaussian distribution
\begin{equation}
\delta (x -x') =lim_{\epsilon \to 0}{\frac{e^{-\frac{\left(x-x'\right)^2}{2 \epsilon ^2}}}{\sqrt{2 \pi } \epsilon }}
 \label{eq:deltaasgaussian}
\end{equation}
\begin{figure}[h!]
\centering
    \includegraphics[width=0.5\textwidth]{dirac.jpg}
\caption{How to think about the Dirac delta function as a form of the Gaussian distribution.}
\end{figure}


\subsection{A physical interpretation}
Enough math! What's the \textit{physical} meaning of $\delta(x-x')?$ Well suppose we have a particle with wavefunction
\begin{equation*}
    \psi(x)=\delta(x-17).
\end{equation*}
It's pretty clear that the particle must be at $x=17$ - the position of the particle is completely determined. In other words, there is no
uncertainty in the position of the particle. What about the momentum? By definition,
\begin{equation*}
    \delta(x-17)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{ik(x-17)}\,dk
\end{equation*}
But this is just a Fourier transform with the momentum-space wavefunction $\phi(k)=1$. The particle has an equal likelihood of having any
momentum! Indeed, this is an extreme case of the uncertainty principle -- completely determining the position of a particle destroys all
knowledge of its momentum (and vice versa). Of course, completely determination is in practice impossible, but it is yet another dramatic
example of quantum weirdness.


\section{Time evolution of the free particle}
Up until now we have only discussed the free particle wavefunction at a fixed time. The time-dependent wavefunction can be written as
\begin{equation}
    \Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\Phi(k,t)e^{ikx}\,dk
    \label{eq:Psi}
\end{equation}
Remember that the dynamical equation in quantum mechanics is the Schr\"{o}dinger equation:
\begin{equation*}
    i\hbar\frac{\partial \Psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi(x,t)}{\partial x^2}+V(x)\Psi(x,t).
\end{equation*}
The left-hand side
\begin{equation*}
        i\hbar\frac{\partial \Psi(x,t)}{\partial t}=\frac{i\hbar}{\sqrt{2\pi}}\int_{-\infty}^\infty\frac{\partial \Phi(k,t)}{\partial t}e^{ikx}\,dk
\end{equation*}
and the right-hand side (note that $V(x)=0$)
\begin{equation*}
    -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi(x,t)}{\partial x^2}=\frac{\hbar^2k^2}{2\sqrt{2}\pi m}\int_{-\infty}^\infty\Phi(k,t)e^{ikx}\,dk
\end{equation*}
must be equal, which yields the differential equation:
\begin{equation*}
    i\hbar\frac{\partial \Phi(k,t)}{\partial t}=\frac{\hbar^2k^2}{2m}\Phi(k,t).
\end{equation*}
Evidently,
\begin{equation}
    \Phi(k,t)=\Phi(k,0)\exp\left(-i\frac{\hbar k^2}{2m}t\right).
    \label{eq:Phi}
\end{equation}
Inserting $\Phi(k,t)$ into Eq.~(\ref{eq:Psi}), we obtain the general, time-dependent wavefunction of the free-particle:
\begin{equation}
    \Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\Phi(x,0)e^{ikx-i\frac{\hbar k^2}{2m}t}\,dk.
    \label{eq:freeparticle}
\end{equation}

So, if you're given the initial position-space wavefunction, $\Psi(x,0)$ of a free particle, simply take its Fourier transform to get
$\Phi(k,0)$, the initial momentum-space wavefunction. Then the particle's position-space wavefunction at any subsequent time $t$ is given by
Eq.~(\ref{eq:freeparticle}).
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